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Give an example of a bounded sequence that has a limit.

Short Answer

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Question: Provide an example of a bounded sequence that has a limit. Answer: The sequence a_n = 1/n is an example of a bounded sequence with a limit. This sequence is bounded, with a lower bound m = 0 and an upper bound M = 1. The limit of the sequence as n goes to infinity is 0.

Step by step solution

01

Choosing a bounded sequence

We can consider the sequence of numbers given by the formula: a_n = \frac{1}{n}. This sequence contains only positive elements since n is a positive integer, and the values will decrease as n increases.
02

Verify that the sequence is bounded

We'll prove that the sequence given by a_n = \frac{1}{n} is bounded. Since a_n is positive for all n, the lower bound m is 0. For the upper bound M, notice that as n increases, the value of \frac{1}{n} decreases. The largest value for a_n occurs when n = 1 and a_1 = 1. Therefore, the upper bound M is 1. Since a_n has both upper and lower bounds, the sequence is bounded.
03

Observe the limit of the sequence

Now, let's determine the limit of the sequence a_n = \frac{1}{n} as n goes to infinity. Intuitively, as n becomes larger, the value of \frac{1}{n} decreases and approaches 0. In mathematical terms, we write this as: \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{1}{n} = 0.
04

Conclusion

We've shown that the sequence a_n = \frac{1}{n} is bounded, with a lower bound m = 0 and an upper bound M = 1. Additionally, we showed that the limit of the sequence as n goes to infinity is 0. Therefore, the sequence a_n = \frac{1}{n} is an example of a bounded sequence that has a limit.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergence of Sequences
Understanding the convergence of sequences is fundamental in the study of calculus. A sequence converges if it approaches a specific value, known as the limit, as the terms progress towards infinity. To say that a sequence \(a_n\) converges to a limit \(L\) is to claim that for every positive number \(\epsilon\), no matter how small, there exists a corresponding natural number \(N\) such that for all \(n \geq N\), the distance between \(a_n\) and \(L\) is less than \(\epsilon\). In simple terms, after some point in the sequence, all terms get arbitrarily close to \(L\) and stay close.

For example, in the given exercise, the sequence \(1/n\) can be shown to converge to 0 by demonstrating that for any small positive \(\epsilon\), there is an \(N\) after which all terms of \(1/n\) are within \(\epsilon\) of 0. This concept is crucial for understanding the behavior of sequences in calculus and has profound implications in areas such as series and functional analysis.
Limits of Sequences
The limit of a sequence is the value that the elements of the sequence approach as the index \(n\) increases without bound. In a more formal sense, if a sequence \(a_n\) has a limit \(L\), then for every \(\epsilon > 0\), there exists an integer \(N\) such that \(\left| a_n - L \right| < \epsilon\) for all \(n \geq N\). If such \(L\) exists, we write \(\lim_{n\to\infty} a_n = L\).

The exercise demonstrates this concept by highlighting that the sequence \(1/n\) has a limit of 0 because, as \(n\) becomes larger, \(1/n\) gets closer and closer to 0. This limit forms the foundational concept for analyzing sequence behaviors and ultimately is integral in the applications of calculus to real-world problems and functions.
Boundedness in Sequences
A sequence is considered to be bounded if there is a real number that serves as an upper bound and another as a lower bound for all terms in the sequence. This means that all elements of the sequence \(a_n\) are contained within the interval \[m, M\], where \(m\) is the lower bound and \(M\) is the upper bound. Mathematically, a sequence \(a_n\) is bounded if \(m \leq a_n \leq M\) for all natural numbers \(n\).

In the provided exercise, the sequence \(a_n = 1/n\) is shown to be bounded because it is always greater than the lower bound of 0 and less than the upper bound of 1. This property of boundedness is a prerequisite for the Bolzano-Weierstrass theorem, which asserts that every bounded sequence has at least one convergent subsequence. Boundedness is a protective characteristic of sequences that ensures they don't 'explode' towards infinity or 'implode' towards negative infinity.
Infinite Series Calculus
Infinite series calculus deals with the sum of infinitely many numbers, arranged in a sequence that may converge to a specific value. An infinite series is the sum of all terms in a sequence \(a_n\), expressed as \(\sum_{n=1}^{\infty} a_n\). If the sequence of partial sums converges, then the series converges to the same limit. When evaluating such series, it is essential to understand whether the sequence of terms \(a_n\) is bounded and if it converges, as these properties can significantly influence the series' behavior.

As an extension of our exercise, if we consider the series formed by the sum of the terms \(1/n\), it is known as the harmonic series, and unlike the sequence, this series diverges, meaning its sum grows without bound. This divergence highlights the subtle but important differences between sequences and series in calculus.

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Most popular questions from this chapter

Consider the number \(0.555555 \ldots,\) which can be viewed as the series \(5 \sum_{k=1}^{\infty} 10^{-k} .\) Evaluate the geometric series to obtain a rational value of \(0.555555 .\) b. Consider the number \(0.54545454 \ldots\), which can be represented by the series \(54 \sum_{k=1}^{\infty} 10^{-2 k} .\) Evaluate the geometric series to obtain a rational value of the number. c. Now generalize parts (a) and (b). Suppose you are given a number with a decimal expansion that repeats in cycles of length \(p,\) say, \(n_{1}, n_{2} \ldots ., n_{p},\) where \(n_{1}, \ldots, n_{p}\) are integers between 0 and \(9 .\) Explain how to use geometric series to obtain a rational form for \(0 . \overline{n_{1}} n_{2} \cdots n_{p}\) d. Try the method of part (c) on the number \(0 . \overline{123456789}=0.123456789123456789 \ldots\) e. Prove that \(0 . \overline{9}=1\)

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=n^{10} \text { and } b_{n}=n^{9} \ln ^{3} n, n \geq 7$$

A fallacy Explain the fallacy in the following argument. Let \(x=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\) and \(y=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots \cdot\) It follows that \(2 y=x+y\) which implies that \(x=y .\) On the other hand, $$ x-y=\underbrace{\left(1-\frac{1}{2}\right)}_{>0}+\underbrace{\left(\frac{1}{3}-\frac{1}{4}\right)}_{>0}+\underbrace{\left(\frac{1}{5}-\frac{1}{6}\right)}_{>0}+\cdots>0 $$ is a sum of positive terms, so \(x>y .\) Therefore, we have shown that \(x=y\) and \(x>y\)

Use Exercise 89 to determine how many terms of each series are needed so that the partial sum is within \(10^{-6}\) of the value of the series (that is, to ensure \(\left|R_{n}\right|<10^{-6}\) ). Functions defined as series Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty}(-1)^{k} x^{k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

Consider the geometric series \(S=\sum_{k=0}^{\infty} r^{k}\) which has the value \(1 /(1-r)\) provided \(|r|<1\). Let \(S_{n}=\sum_{k=0}^{n-1} r^{k}=\frac{1-r^{n}}{1-r}\) be the sum of the first \(n\) terms. The magnitude of the remainder \(R_{n}\) is the error in approximating \(S\) by \(S_{n} .\) Show that $$ R_{n}=S-S_{n}=\frac{r^{n}}{1-r} $$

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